Circular motion with Friction Questions in English

(C) For a particle moving in a horizontal circle on the inner surface of a smooth cone,the forces acting on the particle are its weight $mg$ (downward) and the normal reaction $N$ (perpendicular to the surface of the cone).
The vertical component of the normal reaction balances the weight: $N \cos \theta = mg$.
The horizontal component of the normal reaction provides the necessary centripetal force: $N \sin \theta = \frac{mv^2}{r}$,where $r$ is the radius of the circular path.
Dividing the two equations: $\tan \theta = \frac{v^2}{rg}$.
From the geometry of the cone,the radius $r$ at height $h$ is given by $r = h \tan \theta$.
Substituting $r$ into the equation: $\tan \theta = \frac{v^2}{(h \tan \theta) g}$.
This simplifies to $v^2 = gh \tan^2 \theta$,or $v = \sqrt{gh} \tan \theta$.
Given $h = 10 \ cm = 0.1 \ m$,$g = 10 \ m/s^2$,and assuming $\theta = 45^\circ$ (so $\tan 45^\circ = 1$):
$v = \sqrt{10 \times 0.1} \times 1 = \sqrt{1} = 1 \ m/s$.

(B) The correct option is $B$.
Concept: For safe driving on a level road, the centripetal force required for turning must be provided by the static friction between the tyres and the road. If the centripetal force exceeds the maximum limiting friction, the vehicle will skid.
Mathematically, the condition for safe turning is $\frac{mv^2}{r} \leq \mu mg$.
Given: Velocity $v = 108 \,km/hr = 108 \times \frac{5}{18} \,m/s = 30 \,m/s$, coefficient of friction $\mu = 0.5$, and $g = 10 \,m/s^2$.
To find the minimum radius $r_{\min}$, we use the equality condition: $r_{\min} = \frac{v^2}{\mu g}$.
Substituting the values: $r_{\min} = \frac{30^2}{0.5 \times 10} = \frac{900}{5} = 180 \,m$.

(A) For a car on a banked road,the maximum safe speed $v$ is given by the relation $v = \sqrt{Rg \frac{\mu + \tan \theta}{1 - \mu \tan \theta}}$.
Since the angle of banking $\theta$ and the coefficient of friction $\mu$ remain unchanged,we have $v^2 \propto R$,or $v^2 = C R$,where $C$ is a constant.
Let the initial speed be $v$ and the initial radius be $R = 20 \ m$.
The new speed is $v' = v + 0.10v = 1.1v$.
Using the relation $v^2 = CR$,we have $v'^2 = CR'$,where $R'$ is the new radius.
Dividing the two equations: $\frac{v'^2}{v^2} = \frac{R'}{R}$.
Substituting the values: $(1.1)^2 = \frac{R'}{R} \Rightarrow 1.21 = \frac{R'}{R}$.
Therefore,$R' = 1.21 R = 1.21 \times 20 \ m = 24.2 \ m$.
The increase in the radius of curvature is $\Delta R = R' - R = 24.2 \ m - 20 \ m = 4.2 \ m$.

(D) The angle of banking $\theta$ is given by $\tan \theta = \frac{h}{x}$,where $h$ is the elevation of the outer rail and $x$ is the gauge of the railway line.
Given,the gauge $x = 1 \text{ m}$ and $\tan \theta = \frac{1}{20}$.
Substituting these values into the formula:
$\frac{1}{20} = \frac{h}{1 \text{ m}}$
$h = \frac{1}{20} \text{ m} = 0.05 \text{ m}$.
Converting this to centimeters:
$h = 0.05 \times 100 \text{ cm} = 5 \text{ cm}$.
Thus,the elevation of the outer rail above the inner rail is $5 \text{ cm}$.

(D) For a car moving on a level circular road,the necessary centripetal force is provided by the static friction between the tires and the road.
Let $m$ be the mass of the car,$v$ be the speed,$r$ be the radius of the circular path,and $\mu$ be the coefficient of friction.
The centripetal force required is $F_c = \frac{mv^2}{r}$.
The maximum frictional force available is $f_{max} = \mu N = \mu mg$.
To prevent slipping,we must have $F_c \leq f_{max}$,which implies $\frac{mv^2}{r} \leq \mu mg$.
Thus,the minimum coefficient of friction required is $\mu = \frac{v^2}{rg}$.
Substituting the given values: $v = 12.5 \ m/s$,$r = 20 \ m$,and $g = 9.8 \ m/s^2$:
$\mu = \frac{12.5 \times 12.5}{20 \times 9.8} = \frac{156.25}{196} \approx 0.797$.
Rounding to the nearest provided option,we get $\mu = 0.8$.

(C) Let the natural length of the spring be $L$ and the extension be $x$.
The total radius of the circular path is $R = L + x$.
The restoring force provided by the spring acts as the centripetal force for the circular motion.
$F_{\text{restoring}} = F_{\text{centripetal}}$
$kx = m(L + x)\omega^2$
Given: $m = 200 \text{ g} = 0.2 \text{ kg}$,$k = 12.5 \text{ N/m}$,$\omega = 5 \text{ rad/s}$.
Substituting the values:
$12.5x = 0.2(L + x)(5)^2$
$12.5x = 0.2(L + x)(25)$
$12.5x = 5(L + x)$
$12.5x = 5L + 5x$
$7.5x = 5L$
$\frac{x}{L} = \frac{5}{7.5} = \frac{50}{75} = \frac{2}{3}$
Therefore,the ratio of extension to natural length is $2:3$.

(C) The mass $m$ moves in a horizontal circle of radius $r = \ell \sin \theta$.
The angular frequency is $\omega = 2\pi f = 2\pi \times \frac{1}{\pi} = 2 \text{ rad/s}$.
The forces acting on the mass are tension $T$ (along the string) and weight $mg$ (downwards).
Resolving tension into components: $T \cos \theta = mg$ and $T \sin \theta = m\omega^2 r$.
Substituting $r = \ell \sin \theta$,we get $T \sin \theta = m\omega^2 \ell \sin \theta$,which simplifies to $T = m\omega^2 \ell$.
Substituting $\omega = 2 \text{ rad/s}$,we get $T = m(2)^2 \ell = 4m\ell$.

(B) The maximum speed $v_{max}$ to avoid slipping on a level circular path is given by the condition $v_{max} = \sqrt{\mu_s rg}$.
Squaring both sides,we get $v_{max}^2 = \mu_s rg$.
Given values are $\mu_s = 0.1$,$r = 5 \text{ m}$,and $g = 10 \text{ ms}^{-2}$.
Substituting these values: $v_{max}^2 = 0.1 \times 5 \times 10 = 5 \text{ m}^2\text{s}^{-2}$.
For the vehicle to not slip,the condition is $v^2 \leq v_{max}^2$,which means $v^2 \leq 5 \text{ m}^2\text{s}^{-2}$.
If the square of the actual speed $v^2$ exceeds this value,the vehicle will slip.
Therefore,the person will slip if $v^2 > 5 \text{ m}^2\text{s}^{-2}$.

(D) Given: Mass $m = 1000 \,kg$, Banking angle $\theta = 30^{\circ}$, Coefficient of friction $\mu = 0.2$, $g = 10 \,ms^{-2}$.
Considering the forces acting on the vehicle perpendicular to the inclined surface:
The normal reaction force $N$ balances the component of gravity $mg \cos \theta$ and the component of the frictional force $f \sin \theta$ (assuming the vehicle is at the point of sliding up the incline).
$N = mg \cos \theta + f \sin \theta$
Since $f = \mu N$, we have:
$N = mg \cos \theta + \mu N \sin \theta$
$N(1 - \mu \sin \theta) = mg \cos \theta$
$N = \frac{mg \cos \theta}{1 - \mu \sin \theta}$
Substituting the values:
$N = \frac{1000 \times 10 \times \cos 30^{\circ}}{1 - 0.2 \times \sin 30^{\circ}}$
$N = \frac{10000 \times (\sqrt{3}/2)}{1 - 0.2 \times 0.5} = \frac{5000 \times 1.732}{1 - 0.1} = \frac{8660}{0.9} \approx 9622 \,N$.
However, if we consider the force balance along the vertical axis as per the provided diagram:
$N \cos \theta = mg + f \sin \theta$
$N \cos 30^{\circ} = 10000 + 0.2 N \sin 30^{\circ}$
$N(0.866 - 0.1) = 10000$
$N(0.766) = 10000$
$N \approx 13055 \,N$.

(B) Given: Coefficient of friction,$\mu = 0.5$,Radius of curvature,$r = 16.2 \,m$,Acceleration due to gravity,$g = 10 \,ms^{-2}$.
For a car moving on a flat circular path,the centripetal force is provided by the static friction between the tires and the road.
$f = \frac{mv^2}{r} \leq \mu N = \mu mg$
Therefore,the maximum velocity $v_{max}$ is given by:
$v_{max} = \sqrt{\mu rg}$
$v_{max} = \sqrt{0.5 \times 16.2 \times 10}$
$v_{max} = \sqrt{81} = 9 \,ms^{-1}$
To convert this to $kmh^{-1}$,multiply by $3.6$:
$v_{max} = 9 \times 3.6 = 32.4 \,kmh^{-1}$
Comparing with the options,the correct answer is $32.4 \,kmh^{-1}$.

(B) The necessary centripetal force for a car on a curved unbanked road is provided by the static friction force between the tyres and the road.
For safe turning,the centripetal force must be less than or equal to the maximum static friction force:
$\frac{mV^2}{r} \leq \mu mg$
To find the maximum radius $r$ for a given speed $V$,we use the limiting case:
$\frac{mV^2}{r} = \mu mg$
$r = \frac{V^2}{\mu g}$
Given values are $V = 10 \,ms^{-1}$,$\mu = 0.4$,and $g = 10 \,ms^{-2}$.
Substituting these values into the equation:
$r = \frac{10^2}{0.4 \times 10}$
$r = \frac{100}{4}$
$r = 25 \,m$
Thus,the maximum radius of curvature is $25 \,m$.

(B) The forces acting on the motorcyclist are the gravitational force $mg$ acting downwards and the frictional force $f = \mu N$ acting upwards.
For the motorcyclist to stay in a horizontal circle,the normal force $N$ provided by the wall acts as the centripetal force,so $N = \frac{mv^2}{r}$.
To prevent the motorcyclist from sliding down,the frictional force must balance the gravitational force: $f = mg$.
Substituting $f = \mu N$,we get $\mu N = mg$.
$\mu \left(\frac{mv^2}{r}\right) = mg$.
$\mu = \frac{gr}{v^2}$.
Given $g = 10 \ m \ s^{-2}$,$r = 8.0 \ m$,and $v = 5 \sqrt{5} \ m \ s^{-1}$.
$\mu = \frac{10 \times 8}{(5 \sqrt{5})^2} = \frac{80}{25 \times 5} = \frac{80}{125}$.
$\mu = 0.64$.

(B) The maximum speed $v_{max}$ of a car on a banked road with friction is given by the formula: $v_{max} = \sqrt{rg \left( \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right)}$.
Given $r = 8 \ m$,$\theta = 45^{\circ}$,so $\tan \theta = 1$.
The formula simplifies to $v_{max} = \sqrt{rg \left( \frac{1 + \mu}{1 - \mu} \right)}$.
For the first car with $\mu_1 = 0.5$: $v_1 = \sqrt{8g \left( \frac{1 + 0.5}{1 - 0.5} \right)} = \sqrt{8g \left( \frac{1.5}{0.5} \right)} = \sqrt{8g \times 3} = \sqrt{24g}$.
For the second car with $\mu_2 = 0.4$: $v_2 = \sqrt{8g \left( \frac{1 + 0.4}{1 - 0.4} \right)} = \sqrt{8g \left( \frac{1.4}{0.6} \right)} = \sqrt{8g \times \frac{7}{3}} = \sqrt{\frac{56g}{3}}$.
The ratio $v_1 : v_2 = \sqrt{24g} : \sqrt{\frac{56g}{3}} = \sqrt{24} : \sqrt{\frac{56}{3}} = \sqrt{72} : \sqrt{56} = \sqrt{9 \times 8} : \sqrt{7 \times 8} = 3\sqrt{8} : \sqrt{7\times 8} = 3 : \sqrt{7} = \sqrt{9} : \sqrt{7}$.

(C) For the block $P$ not to slide down,the frictional force $f$ must balance the weight $mg$,so $f = mg$.
Since $f \leq \mu N$,we have $mg \leq \mu N$,where $N$ is the normal force provided by the wall.
The normal force $N$ is the centripetal force,$N = m \omega^2 R$.
Thus,for the limiting case of not sliding,$mg = \mu m \omega^2 R$,which implies $\omega^2 R = \text{constant}$ (assuming $\mu$ and $m$ are the same for all cases).
Therefore,$\omega^2 R = C$ (a constant),which means $\omega \propto \frac{1}{\sqrt{R}}$.
Given the radii $R_A < R_B < R_C$,it follows that $\frac{1}{\sqrt{R_A}} > \frac{1}{\sqrt{R_B}} > \frac{1}{\sqrt{R_C}}$.
Consequently,the angular velocities must satisfy $\omega_A > \omega_B > \omega_C$.

(D) The car experiences two accelerations: tangential acceleration $a_t = 3 \ m \ s^{-2}$ and centripetal acceleration $a_c = \frac{v^2}{R}$.
The total acceleration is $a = \sqrt{a_t^2 + a_c^2} = \sqrt{3^2 + (\frac{v^2}{16})^2}$.
The frictional force provides the necessary centripetal and tangential force,so the maximum frictional force $f_{max} = \mu N = \mu mg$ must be greater than or equal to the net force $F = ma$.
Thus,$\mu mg \geq m \sqrt{a_t^2 + a_c^2}$.
Squaring both sides: $(\mu g)^2 \geq a_t^2 + (\frac{v^2}{R})^2$.
Substituting the values: $(0.5 \times 10)^2 \geq 3^2 + (\frac{v^2}{16})^2$.
$25 \geq 9 + \frac{v^4}{256}$.
$16 \geq \frac{v^4}{256}$.
$v^4 \leq 16 \times 256 = 4096$.
$v \leq (4096)^{1/4} = 8 \ m \ s^{-1}$.

(D) The maximum speed $V_{\max}$ on a banked road with friction is given by $V_{\max} = \sqrt{\frac{rg(\mu + \tan \theta)}{1 - \mu \tan \theta}}$.
The optimum speed $V_o$ (where no friction is required) is given by $V_o = \sqrt{rg \tan \theta}$.
Given $\theta = 45^{\circ}$ and $V_{\max} = 2V_o$.
Substituting the expressions: $\sqrt{\frac{rg(\mu + \tan \theta)}{1 - \mu \tan \theta}} = 2 \sqrt{rg \tan \theta}$.
Squaring both sides: $\frac{\mu + \tan \theta}{1 - \mu \tan \theta} = 4 \tan \theta$.
Since $\tan 45^{\circ} = 1$,we have $\frac{\mu + 1}{1 - \mu} = 4(1)$.
$\mu + 1 = 4 - 4\mu$.
$5\mu = 3$.
$\mu = \frac{3}{5} = 0.6$.

(C) For the child not to slip,the required centripetal force must be provided by the static friction force.
At the point of slipping,the centripetal force equals the maximum static friction force:
$m \omega^2 r = f_{s, \text{max}}$
Since $f_{s, \text{max}} = \mu N$ and the normal force $N = mg$,we have:
$m \omega^2 r = \mu mg$
$
\omega^2 = \frac{\mu g}{r}
$
Given values: $\mu = 0.8$,$g = 10 \ m/s^2$,and $r = 2 \ m$.
Substituting these values:
$
\omega^2 = \frac{0.8 \times 10}{2} = \frac{8}{2} = 4
$
$
\omega = \sqrt{4} = 2 \ rad/s
$
Thus,the maximum angular velocity is $2 \ rad/s$.

(C) The maximum speed $v_{max}$ on a banked road with friction is given by the formula: $v_{max} = \sqrt{rg \left( \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right)}$.
Given: Radius $r = 75 \ m$,angle $\tan \theta = 0.2$,coefficient of friction $\mu = 0.1$,and acceleration due to gravity $g = 10 \ m/s^2$.
Substituting the values:
$v_{max} = \sqrt{75 \times 10 \times \left( \frac{0.2 + 0.1}{1 - (0.1 \times 0.2)} \right)}$
$v_{max} = \sqrt{750 \times \left( \frac{0.3}{1 - 0.02} \right)}$
$v_{max} = \sqrt{750 \times \left( \frac{0.3}{0.98} \right)}$
$v_{max} = \sqrt{750 \times 0.3061} \approx \sqrt{229.57} \approx 15.15 \ m/s$.
Rounding to the nearest option,the maximum permissible speed is $15 \ m/s$.

(D) When a cyclist negotiates a circular turn, they lean at an angle $\theta$ with the vertical. The forces acting on the cyclist are the weight $mg$ acting downwards and the normal reaction $N$ from the ground.
Resolving the normal reaction $N$ into two components:
$N \cos \theta = mg$ (vertical component balancing the weight) ... $(i)$
$N \sin \theta = \frac{mv^2}{R}$ (horizontal component providing the necessary centripetal force) ... (ii)
Dividing equation (ii) by equation $(i)$, we get:
$\tan \theta = \frac{v^2}{Rg}$
Given $\theta = 30^{\circ}$, $R = 20 \sqrt{3} \,m$, and taking $g = 10 \,m/s^2$:
$\tan 30^{\circ} = \frac{v^2}{20 \sqrt{3} \times 10}$
$\frac{1}{\sqrt{3}} = \frac{v^2}{200 \sqrt{3}}$
$v^2 = \frac{200 \sqrt{3}}{\sqrt{3}} = 200$
$v = \sqrt{200} = 10 \sqrt{2} \,m/s$.
Wait, re-evaluating the standard formula: $\tan \theta = \frac{v^2}{Rg}$. If the angle is with the vertical, $\tan 30^{\circ} = \frac{v^2}{Rg}$.
$v^2 = Rg \tan 30^{\circ} = (20 \sqrt{3}) \times 10 \times \frac{1}{\sqrt{3}} = 200$.
$v = \sqrt{200} = 10 \sqrt{2} \,m/s$.
Given the options, if the angle $30^{\circ}$ is with the horizontal, then $\tan 60^{\circ} = \frac{v^2}{Rg}$.
$v^2 = (20 \sqrt{3}) \times 10 \times \sqrt{3} = 200 \times 3 = 600$.
$v = \sqrt{600} = 10 \sqrt{6} \,m/s$. Thus, option $D$ is correct.

(C) Let the body leave the surface of the hemisphere at point $P$.
At point $P$,let the radius vector of the body make an angle $\theta$ with the vertical.
The forces acting on the body at point $P$ are the gravitational force $mg$ (downwards) and the normal reaction $R$ (outwards).
The component of gravitational force towards the center is $mg \cos \theta$.
The net centripetal force is provided by the difference between the radial component of gravity and the normal reaction:
$mg \cos \theta - R = \frac{mv^2}{r}$
When the body leaves the surface,the normal reaction $R$ becomes zero.
Therefore,$mg \cos \theta = \frac{mv^2}{r}$
Substituting the given values $v = 5 \text{ m/s}$,$r = 5 \text{ m}$,and $g = 10 \text{ m/s}^2$:
$\cos \theta = \frac{v^2}{rg} = \frac{5^2}{5 \times 10} = \frac{25}{50} = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ}$.

(D) For the body of mass $M$ to remain stuck to the inner wall of the rotating drum,the normal force $N$ exerted by the wall provides the necessary centripetal force:
$N = M \omega^2 R$
The frictional force $f$ acts vertically upwards to balance the gravitational force $Mg$ acting downwards:
$f = Mg$
Since the body is on the verge of slipping,the frictional force is at its maximum value,given by $f = \mu N$:
$Mg = \mu N$
Substituting the expression for $N$ into the friction equation:
$Mg = \mu (M \omega^2 R)$
Dividing both sides by $M$ and solving for $\omega$:
$g = \mu \omega^2 R$
$\omega^2 = \frac{g}{\mu R}$
$\omega = \sqrt{\frac{g}{\mu R}}$

(A) For the mass to remain stuck to the wall,the force of static friction $(f)$ must balance the gravitational force $(mg)$: $f = mg$.
Since $f = \mu N$,where $N$ is the normal force,we have $\mu N = mg$.
The normal force is provided by the centripetal force: $N = m\omega^2R$.
Substituting $N$ into the friction equation: $\mu (m\omega^2R) = mg$.
Solving for $\mu$: $\mu = \frac{g}{\omega^2R}$.
Given $g = 10 \ m/s^2$,$\omega = 5 \ rad/s$,and $R = 4 \ m$:
$\mu = \frac{10}{5^2 \times 4} = \frac{10}{25 \times 4} = \frac{10}{100} = 0.1$.